RELATION BETWEEN SOLVABILITY OF SOME MULTIVARIATE INTERPOLATION PROBLEMS AND THE VARIETY OF SUBSPACE ARRANGEMENT. The subspace arrangement is=where are linear subspaces of .

=

}=d-s

Algebraic geometry studies the property

Birkhoff Interpolation(in one variable)

George David Birkhoff

Birkhoff interpolation is an extension of Hermite interpolation. It involves matching of values and derivatives of a function at certain points without the requirement that the derivatives are consecutive.

Example: find a polynomial 𝑓 (𝑧1 )=𝑎1 , 𝑓 ′ (𝑧1)=𝑎2 ,𝑓 ′ (𝑧 2 )=𝑎3𝑓 ′ ′ ′ (𝑧2 )=𝑎4

If this equations have unique solution for all distinct and all the problem ( is regular.

=𝞅(, )det

) :𝞅(, )}

¿6 𝑧 2−6 𝑧1

?⊆𝓗={¿

Example: find a polynomial

𝑓 (𝑧1 )=𝑎1 ,𝑓 ′ ′ (𝑧 1)=𝑎2 , , 𝑓 ′ ′ ′ (𝑧2 )=𝑎4

the problem ( is regular, completely regular.

) :𝞅(, )=0}=

=𝞅 ( 𝑧 1 ,  𝑧2 )  =12det

=

Birkhoff Interpolation problem:

Isaac SchoenbergGeorge Pólya

, where ; is regular if for any set of k district points and any f there exists unique p such that )=) for all k and all j=1,…,k.

The problem , is regular if1) k=n+1, ={0}, Lagrange interpolation2) For k>0, k , Hermite interpolation3) , ( = ).

ℂ  𝑑 ,𝑑>1

George G. Lorentz

The same is true in real case

=(, =(,

𝞅(, )= =𝞅(, , )

If 𝞅 =3

; dim =3 if 𝞅If the scheme is regular then

: }dim =2

If 𝞅 and then the determinant contains two identical rows (columns), hence

Carl de Boor

Amos Ron

Given a subspace a collection of subspaces , a set of points and a function f ] we want to find a polynomial p such that

()f()= ()f()for all qThe scheme (, is regular if the problem has a unique solutions for all f ] and all distinct , it is completely regular if it has unique solution for all .

Birkhoff Interpolation (in several variables)

Birkhoff Interpolation (in several variables)

𝞅(,…, ) is a polynomial in

If 𝞅 =dk-1=

=dim =max{dim}=kd-d

Theorem: If (, is regular then

Rong-Qing Jia A. Sharma

Conjecture: it is true in the real case

False

HAAR SUBSPACES AND HAAR COVERINGS

Alfréd Haar

Definition: H=span {

the determinant

[x].

For d>1, n>1 there are no n-dimensional Haar subspaces in or even in C()

J. C. Mairhuber in real case and I. Schoenberg in the complex case;.

(The Lagrange interpolation problem is well posed)

or from the previous discussion: (,{0},{0},…,{0})

Definition: A family of n-dimensional subspaces {

the Lagrange interpolation problem is well posed in one of these spaces.

=span {

{

Δ𝑘 (𝑧1 ,…,   𝑧𝑛 )=𝑑𝑒𝑡 (h1(𝑘)(𝑧 𝑗))≠0Question: What is the minimal number s:= of n-dimensional subspaces ?

Conjecture: :

Kyungyong Lee

𝐻1=𝑠𝑝𝑎𝑛 {1 , 𝑥 } ,𝐻2=𝑠𝑝𝑎𝑛 {1 , 𝑦 } 𝑥

𝑦

Theorem (Stefan Tohaneanu& B.S.):

It remains to show that no two subspaces can do the job.

,

The three spaces

=span {

=0

I want show that there are three distinct points

), ),

𝓗≔ {(0,0 , a ,b ) }∪ {(a ,b ,0,0)∪{(a ,b ,a , b)}

𝒵 (Δ1 ,Δ2 )=𝓗?

dim =2=dim

Not every two-dimensional variety in can be formed as a set of common zeroes of two polynomials. The ones that can are called (set theoretic) complete intersections.As luck would have it, is not a complete intersection:

is not connected… A very deep theorem states that a two dimensional complete intersection in can not be disconnected by removing just one point

Robin HartshorneAlexander Grothendieck

={(): =0}

𝓗≔ {(0,0 , a , b ) }∪ {(a ,b ,0,0)∪{(a ,b ,a , b)}(0,0,1,1)

(-1,-1,0,0)

(0,0,0,0)𝓗

=3

=

,

The family of D-invariant subspaces spanned by monomials form a finite Haar covering. There are too many of them: for n=4 in 2 dimensions there are five:

𝑦 ¿ ¿1 𝑥 ¿

¿¿1 𝑥 𝑥2 𝑥3 ¿𝑦 2 ¿ ¿𝑦 ¿ ¿1 𝑥 ¿

¿

𝑦 𝑥𝑦 ¿ ¿1 𝑥 ¿ ¿𝑦 3 ¿ ¿𝑦2 ¿ ¿𝑦 ¿ ¿1 ¿ ¿

Yang tableaux

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